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September 30, 2014

September 30, 2014

Posted by **Jim** on Wednesday, November 2, 2011 at 1:07pm.

- Grade 11 Functions -
**Steve**, Wednesday, November 2, 2011 at 1:34pmassuming the ball's trajectory to be a parabola,

y = ax^2 + bx + c

Plug in the three points and solve for y=0, because that;'s how high the ball would have been when it hit the ground.

1 = a*0 + b*0 + c

so, c=1

y = ax^2 + bx + 1

2.5 = 4a + 2b

18 = 17956a + 134b

multiply the first by 67 and subtract from the second, and we have

a = -0.00845

b = 1.2669

s, we have

y = -.00845x^2 + 1.2669x + 1

When y=0, x = 150.714

Must have hit into the stands about halfway to the outfield fence.

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